If the value of the integral $I = \int_{-\pi}^{\pi} \max(\sin x, \tan x)dx$ is equal to $\ln k$, then the value of $k^2$ is equal to
Step-by-Step Solution
Key Concept: Identifying the dominant function in a max expression and integrating correctly with proper bounds
For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$, we have $\tan x > \sin x$ and $\cos x < 1$. The maximum of $(\tan x \sin x)$ occurs when $\tan x \sin x = \tan x$. Thus $\max(\tan x, \sin x) = \tan x$. Therefore $\int \tan x\,dx = [-\ln|\cos x|]^{\pi/4}_{\pi/2} = (-\ln|\cos x|) = -(\ln(1) - \ln(\frac{\sqrt{2}}{2})) = \frac{1}{2}\ln 2$.
Correct Answer: 2